Why is $x^5+x^4-x-1=(x-1)(x+1)^2(x^2+1)$?

Question

How can I prove $x^5+x^4-x-1=(x-1)(x+1)^2(x^2+1)$ by polynomial division? I see that $x=1$ is a zero of the denominator, but how do I get the other factors? Our professor said polynomial division, but I don't see it.

BTW, I know how to get the solution with the binomial formula: \begin{align*} x^5+x^4-x-1 & = (x^5 - x) + (x^4 - 1)\\ & = x(x^4 - 1) + (x^4 - 1)\\ & = (x + 1)(x^4 - 1)\\ & = (x+1)(x^2 + 1)(x^2 - 1)\\ & =(x+1)^2(x^2 + 1)(x - 1)\\ & = \cdots \end{align*}

But I want to know another way to obtain the result.

Answer 1

$\frac{d}{dx}(x^5+x^4-x-1)|_{x=-1}=5(-1)^4+4(-1)^3-1=0$, but $\frac{d^2}{dx^2}(x^5+x^4-x-1)|_{x=-1}=20(-1)^3+12(-1)^2-1\neq 0$, this implies that $x=-1$ is a root of multiplicity two, so $(x-1)^2$ divides $x^5+x^4-x-1$, and as you point out $x=1$ is a root, then $(x+1)(x-1)^2=x^3+x^2-x-1$ divides $x^5+x^4-x-1$. If you use polynomial division $x^5+x^4-x-1\div x^3+x^2-x-1$ easily, you get that the remaining factor is $x^2+1$

Answer 2

Another approach $$ \begin{align} \overbrace{\ x^5+x^4\ }^{(x+1)x^4}\overbrace{\,-x-1\ \vphantom{x^4}}^{(x+1)(-1)} &=(x+1)\left(x^4-1\right)\tag{1a}\\ &=(x+1)\left(x^2-1\right)\left(x^2+1\right)\tag{1b}\\ &=(x+1)(x+1)(x-1)\left(x^2+1\right)\tag{1c}\\ &=(x-1)(x+1)^2\left(x^2+1\right)\tag{1d} \end{align} $$ Explanation:
$\text{(1a):}$ Note that $x+1$ is a factor
$\text{(1b):}$ $u^2-1=(u-1)(u+1)$ with $u=x^2$
$\text{(1c):}$ $u^2-1=(u-1)(u+1)$ with $u=x$
$\text{(1d):}$ reorganize

Answer 3

If we want proceed by long division for the root $x=1$ we obtain

$$\begin{array}{c|rrrrrr}& x^5 & x^4 & x^3 & x^2 & x^1 & x^0 \\ & 1 & 1 & 0 & 0& -1 & -1\\ {\color{red}1} & \downarrow & 1 & 2 & 2 &2 &1\\ \hline & 1 & 2 & 2 & 2 &1&|\phantom{-} {\color{blue}0} \end{array}$$

that is

$$x^5+x^4-x-1=(x-1)(x^4+2x^3+2x^2+2x+1)$$

and for $x^4+2x^3+2x^2+2x+1$ for the root $x=-1$ we obtain

$$\begin{array}{c|rrrrr}& x^4 & x^3 & x^2 & x^1 & x^0 \\ & 1 & 2 & 2& 2 & 1\\ {\color{red}{-1}} & \downarrow & -1 & -1 &-1 &-1\\ \hline & 1 & 1 & 1 &1&|\phantom{-} {\color{blue}0} \end{array}$$

that is

$$x^4+2x^3+2x^2+2x+1=(x+1)(x^3+x^2+x+1)$$

and for $x^3+x^2+x+1$ for the root $x=-1$ we obtain

$$\begin{array}{c|rrrr}& x^3 & x^2 & x^1 & x^0 \\ & 1 & 1& 1 & 1\\ {\color{red}{-1}} & \downarrow & -1 &0 &-1\\ \hline & 1 & 0 &1&|\phantom{-} {\color{blue}0} \end{array}$$

that is

$$x^3+x^2+x+1=(x+1)(x^2+1)$$

and finally

$$x^5+x^4-x-1=(x-1)(x+1)^2(x^2+1)$$

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More directly recognizing by inspection $x=1$ and $x=-1$ as roots

$$\begin{array}{rrrrrr|ll} x^5 & x^4 & & & -x & -1 & x^2 - 1 \\ -x^5 & & x^3 & & & & x^3 +x^2+x+1 \\ \hline & x^4 & x^3 & & -x &-1 \\ & \phantom{-}-x^4 & &x^2 & & & \\ \hline & & x^3 & x^2&-x&-1 \\ & &-x^3 & & x& \\ \hline & & & x^2& &-1 \\ & & &-x^2 & &+1 \\ \hline & & & & & \end{array}$$

that is

$$x^5+x^4-x-1=(x+1)(x-1)(x^3+x^2+x+1)$$

and since $(x-1)(x^3+x^2+x+1)=x^4-1=(x^2+1)(x^2-1)$

we have

$$x^5+x^4-x-1=(x+1)^2(x-1)(x^2+1)$$

Answer 4

Yet another way to solve it, without polynomial divisions.

• By inspection, two of the roots are $\,1\,$ and $\,-1\,$.

• Also by inspection, the derivative $\,5 x^4 + 4 x^3 - 1\,$ has $\,-1\,$ as a root, so $\,-1\,$ is a double root.

• By Vieta's relations, the remaining two roots $\,a,b\,$ must satisfy $\, 1 + 2 \cdot (-1) + a + b = -1\,$ $\iff a+b=0\,$ and $\, 1 \cdot (-1)^2 \cdot ab = 1 \iff ab = 1\,$, so the quadratic factor is $\,x^2 + 1\,$.